Question

If $x$ is a cube root of unity other than 1 , then
${\left(x+\frac{1}{x}\right)}^2+{\left(x^2+\frac{1}{x^2}\right)}^2+\dots +{\left(x^{12}+\frac{1}{x^{12}}\right)}^2=$

• A. 12
• B. 64
• C. 24
• D. 0

Answer 1

Answer: (C)

Given, $x$ is a cube root of unity other than 1 i.e. $x=\omega$ or ${\omega }^2$
Now, ${\left(x+\frac{1}{x}\right)}^2+{\left(x^2+\frac{1}{x^2}\right)}^2+\dots +{\left(x^{12}+\frac{1}{x^{12}}\right)}^2$
$={\left(\omega +\frac{1}{\omega }\right)}^2+{\left({\omega }^2+\frac{1}{{\omega }^2}\right)}^2+\dots +{\left({\omega }^{12}+\frac{1}{{\omega }^{12}}\right)}^2$
$={\left(\omega +\frac{1}{\omega }\right)}^2+{\left({\omega }^2+\frac{1}{{\omega }^2}\right)}^2+\dots +{\left({\omega }^{11}+\frac{1}{{\omega }^{11}}\right)}^2$
$={\left(\omega +{\omega }^2\right)}^2+{\left({\omega }^2+\omega \right)}^2+(1+1{)}^2+{\left(\omega +{\omega }^2\right)}^2$
$+{\left({\omega }^2+\omega \right)}^2+(1+1{)}^2+{\left(\omega +{\omega }^2\right)}^2$
$=8{\left(\omega +{\omega }^2\right)}^2+4(1+1{)}^2$
$=8(-1{)}^2+4(2{)}^2=8+16=24$